Amaya is standing 30 ft from a volleyball not. The not is 8 It high. Amaya serves the ball.The path of the ball is modeled by the equation y=-0.02(x-18)^2+12 where x is the ball's horizontal distance in foot from Amaya's position and y is the distance in feet from the ground to the ball a. How far away is the ball from Amaya when it is at its maximum height?Explain b. Describe how you would find the ball's hoight when it crosses the not at x=30 a. How far away is the ball from Amaya when it is at its maximum height?Explain The ball is square It away from Amaya when it is at its maximum hoight. Use the square of the vertex

a. To find the horizontal distance of the ball from Amaya when it is at its maximum height, we need to find the x-coordinate of the vertex of the parabola.<br /><br />The equation of the path of the ball is given as:<br />$y = -0.02(x-18)^2 + 12$<br /><br />The vertex form of a quadratic equation is:<br />$y = a(x-h)^2 + k$<br /><br />Comparing the given equation with the vertex form, we can see that the vertex is at (18, 12).<br /><br />Therefore, the ball is 18 feet away from Amaya when it is at its maximum height.<br /><br />b. To find the height of the ball when it crosses the net at $x=30$, we need to substitute $x=30$ into the equation of the path of the ball.<br /><br />Substituting $x=30$ into the equation:<br />$y = -0.02(30-18)^2 + 12$<br />$y = -0.02(12)^2 + 12$<br />$y = -0.02(144) + 12$<br />$y = -2.88 + 12$<br />$y = 9.12$<br /><br />Therefore, the height of the ball when it crosses the net at $x=30$ is 9.12 feet.